Problem 4 - A Complete Resource Book in Chemistry for JEE Main

Question

The oxidation number of \(\mathrm{V}\) in \(\mathrm{Cs}_{4} \mathrm{Na}\left(\mathrm{HV}_{10} \mathrm{O}_{28}\right)\) is (a) \(+2\) (b) \(+5\) (c) \(-2\) (d) \(+3\)

Step-by-Step Solution

Answer

The oxidation state of \(\mathrm{V}\) is \(+5\).

Step 1: Identify Known Oxidation States

Understand that the oxidation number of cesium (Cs) is +1, sodium (Na) is +1, and oxygen (O) is typically -2 in compounds. We will use these to find the oxidation state of vanadium (V).

Step 2: Set Up the Equation for the Compound

The overall compound is electrically neutral. Therefore, we can set up the equation for the oxidation states: \(4(+1) + 1(+1) + 10(x) + 28(-2) = 0\), where \(x\) is the oxidation state of vanadium.

Step 3: Simplify the Equation

Calculate the known parts: \(4(+1) = +4\) for Cs, \(+1\) for Na, and \(28(-2) = -56\) for O. Substitute these into the equation: \(+4 + 1 + 10x - 56 = 0\).

Step 4: Solve for Vanadium

Combine and simplify the numbers in the equation: \(+5 - 56 + 10x = 0\), which simplifies to \(-51 + 10x = 0\).

Step 5: Determine the Oxidation State

Solve for \(x\) by adding 51 to both sides: \(10x = 51\). Then divide by 10: \(x = +5.1\). Round to the nearest whole number for common oxidation states to find \(x = +5\).

Key Concepts

Oxidation State

In chemistry, the concept of oxidation states, also known as oxidation numbers, is crucial for understanding redox reactions and the electron distribution within a compound. An oxidation state refers to the formal charge an atom would have if all bonds to atoms of different elements were 100% ionic. This is especially important when dealing with transition metals, like vanadium, which can have multiple oxidation states.
An oxidation state can be determined in a compound by using a systematic approach:

• Identify known oxidation states of common elements. For example, oxygen typically has an oxidation state of -2 in compounds.
• Sum the total oxidation states in a compound and set it equal to the molecule's net charge. For neutral compounds, this sum should be zero.
• Solve for the unknown oxidation state. In our case, this would be for vanadium.

This balance allows us to determine how electrons are gained, lost, or shared, providing insight into the compound's chemical behavior.

Vanadium

Vanadium is a transition metal with varied oxidation states, making it particularly interesting in the study of oxidation numbers. It commonly exhibits oxidation states ranging from +2 to +5.
In compounds like Cs4Na(HV10O28), vanadium's oxidation state can significantly affect the compound's properties and reactivity. Determining vanadium's oxidation state involves:

• Account for the oxidation states of other elements in the compound, such as cesium +1, sodium +1, and oxygen -2.
• Balance the overall charge to maintain neutrality, particularly for neutral compounds like Cs4Na(HV10O28).
• Use algebra to determine vanadium's oxidation state. For example, this process involves balancing the equation derived from the sum of all contributing elements.

Vanadium's diverse oxidation states allow it to engage in various chemical reactions, often making it a component in industrial catalysts and other significant chemical processes.

Neutral Compound

A neutral compound is a chemical compound that has no net electric charge. This means the total positive charge of the compound's cations equals the total negative charge from its anions.
Working with neutral compounds requires calculating oxidation states to ensure the charges are balanced. Here's how this applies to our specific compound Cs4Na(HV10O28):

• Calculate the individual charges of constituent atoms. Example: cesium (Cs) and sodium (Na) each with +1, and oxygen (O) with -2.
• Use these known oxidation states to set up an equation that sums up to zero, since the compound is neutral.
• This equation helps us find unknown oxidation states, such as that of vanadium in this compound.

Understanding neutral compounds' charge balance helps in predicting how compounds will interact chemically and physically, offering insight into their stability and reactivity.